Question

Let \( f : \mathbb{R}^2 \to \mathbb{R} \) (where \( \mathbb{R} \) is the set of real numbers) be defined by \[ f(x, y) = \begin{cases} \frac{(x - y)^3}{x^2 + y^2}, & (x, y) \neq (0, 0) \\ 0, & (x, y) = (0, 0) \end{cases} \] If \( f_x(0, 0) \) and \( f_y(0, 0) \) denote partial derivatives of \( f \) with respect to \( x \) and \( y \) at the point \( (0, 0) \), respectively, then \( f_x(0, 0) \) and \( f_y(0, 0) \), respectively, are:

• A. 1 and 1
• B. 1 and 2
• C. 1 and -1
• D. 2 and 1

Hint:

When computing partial derivatives, especially when dealing with indeterminate forms, be sure to use L'Hopital's Rule or direct substitution to simplify the expressions.

Answer 1

The Correct Option is C

To solve this problem, we need to compute the partial derivatives of the function \( f(x, y) \) at the point \( (0, 0) \). Step 1: Compute \( f_x(0, 0) \)
The partial derivative of \( f \) with respect to \( x \) is given by: \[ f_x(x, y) = \frac{\partial}{\partial x} \left( \frac{(x - y)^3}{x^2 + y^2} \right) \] We will first apply the quotient rule and then substitute \( (x, y) = (0, 0) \). Using L'Hopital's Rule or direct substitution, we can compute this derivative and find that: \[ f_x(0, 0) = 1 \] Step 2: Compute \( f_y(0, 0) \)
Similarly, the partial derivative of \( f \) with respect to \( y \) is: \[ f_y(x, y) = \frac{\partial}{\partial y} \left( \frac{(x - y)^3}{x^2 + y^2} \right) \] Again, applying the quotient rule and substituting \( (x, y) = (0, 0) \), we find: \[ f_y(0, 0) = -1 \] Step 3: Conclusion
Thus, the correct answer is (C) 1 and -1, where \( f_x(0, 0) = 1 \) and \( f_y(0, 0) = -1 \).